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If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?
(1) a^n = 64
(2) n=6
(1) a^n = 64
(2) n=6
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subhajeet
Prime factorization would be the best way to attack such kind of questions.
Given: \(a^n*k=8!=2^7*3^2*5*7\).
Question: \(a=?\)
(1) \(a^n=64=2^6=4^3=8^2\), so \(a\) could be 2, 4, or 8. Not sufficient.
(2) \(n=6\) --> the only integer (more than 1), which is a factor of 8!, and has the power of 6 (at least) is 2, hence \(a=2\). Sufficient.
Answer: B.
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This one is a fun question (a good practice of our understanding of factors)!
Analyze the given first before delving into the statements.
\(8*6*7*5*4*3*2*1 = a^n * R\)
Statement (1): a^n = 64
\(8*6*7*5*4*3*2*1 = 64 * R\)
64 could be 2^6 or 8^2.
In 8!, it contains at least 6 factors of 2. In 8!, it also contains 2 factors of 8. Thus, a could be 2 or 8.
Thus, INSUFFICIENT.
Statement (2): n = 6
Let us analyze 8! = 8*7*6*5*4*3*2*1. How many prime factors have at least 6 factors in 8!.
Let us start with a = 2. YES!
Let us then try a=3. NO!
We are certain that 2 has the most number of factors in 8! and it has at least 6. SUFFICIENT.
a = 2
AnsweR: B
Analyze the given first before delving into the statements.
\(8*6*7*5*4*3*2*1 = a^n * R\)
Statement (1): a^n = 64
\(8*6*7*5*4*3*2*1 = 64 * R\)
64 could be 2^6 or 8^2.
In 8!, it contains at least 6 factors of 2. In 8!, it also contains 2 factors of 8. Thus, a could be 2 or 8.
Thus, INSUFFICIENT.
Statement (2): n = 6
Let us analyze 8! = 8*7*6*5*4*3*2*1. How many prime factors have at least 6 factors in 8!.
Let us start with a = 2. YES!
Let us then try a=3. NO!
We are certain that 2 has the most number of factors in 8! and it has at least 6. SUFFICIENT.
a = 2
AnsweR: B
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Expert reply
Video solution from Quant Reasoning starts at 25:41
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
General Discussion
D
the product of the first 8 positive integers is
\(P=1*2*3*4*5*6*7*8=2*3*2^2*5*(3*2)*7*2^3=2^6*3^2*5*7\)
\(2^6=64\)
1. only \(2^6\) works. suff.
2. only \(2^6\) works. suff.
the product of the first 8 positive integers is
\(P=1*2*3*4*5*6*7*8=2*3*2^2*5*(3*2)*7*2^3=2^6*3^2*5*7\)
\(2^6=64\)
1. only \(2^6\) works. suff.
2. only \(2^6\) works. suff.
dc123
Such questions try to trick you with an 'obvious c' i.e. you can very easily get the answer using both the statements together. The trick is generally to find the one statement which alone is sufficient (more often that not, there will be one statement which is sufficient in such questions)
In this question we know that a and n are positive integers greater than 1.
Using both the statements together, we can easily find 'a'. But there is a catch.
Let us review each statement independently.
Given: product of the first 8 positive integers is a multiple of a^n
\(1*2*3*4*5*6*7*8 = a^n * k\) (where k is a positive integer)
What does this imply? It implies that every factor of a^n has to be a factor of \(1*2*3*4*5*6*7*8\) as well.
1) \(a^n = 64\)
64 will be a factor of \(1*2*3*4*5*6*7*8\), of course. But can we find 'a' now?
\(2^6 = 64; 4^3 = 64; 8^2 = 64\)
a can take any value 2/4/8. Not sufficient.
2. n = 6
'a' can be 2 since 2^6 = 64 which is a factor of \(1*2*3*4*5*6*7*8 = 64*3*5*6*7\)
Can 'a' be 3?
Is 3^6 a factor of 1*2*3*4*5*6*7*8? No, it has only two 3's. It doesn't have six 3's.
Can 'a' be 4?
Is 4^6 a factor of 1*2*3*4*5*6*7*8? No, it has only three 4's.
The same thing will be true for any number greater than 3 too.
Since 'n' is fixed here i.e. 6, 'a' can take only one value i.e. 2
Hence stmnt 2 alone is sufficient. 'a' can only be 2.
Answer (B)
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i have a question here, what if n = 2 or 3? would the answer be E? or C?
supreetb
The answer would be C:
(1)+(2) a^n = 64 and n=2 --> a^2=64 --> a=8 (discard a=-8 since we know that a is a positive integer). Sufficient.
Hope it's clear.
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I don't think this question is written well. I get this part:
So there are two integers, a and n. They are both greater than 1, Got it. The problem is I don't get this part:
The product of the first 8 positive integers OF WHAT? What are we multiplying? a and n together? Is there a set we are looking at?
Quote:
So there are two integers, a and n. They are both greater than 1, Got it. The problem is I don't get this part:
Quote:
The product of the first 8 positive integers OF WHAT? What are we multiplying? a and n together? Is there a set we are looking at?
joondez
The product of the first 8 positive integers is 1*2*3*4*5*6*7*8. Check completer solution here: https://gmatclub.com/forum/if-the-integ ... l#p1084330
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Bunuel
In that case the question should be:
Quote:
joondez
The question is correct as it is. The first 8 positive integers are 1, 2, 3, 4, 5, 6, 7, and 8. What does "of all real numbers" has to do here? Also, notice that this is an official question from GMAT Prep, so it's as correct as it gets.
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When the question asked for the first 8 positive numbers, it was clearly describing a set. However, it wasn't clear to me what set they were describing. I spent time wondering if they were describing a set of "a" multiple, "n" multiples, or "a^n" multiples. There are many other GMAT questions were the wording is extremely specific but for this question you had to assume that we were looking at all real numbers, which is not always the case for other GMAT questions. So to me this is a very poorly worded question. You are very good at GMAT questions so you may have been able to recognize the question pattern right away, but for someone who is unable to make assumptions based on the question maker's mind, this question was not solvable. And the fact that this is a GMATPrep question does not make it infallible, in fact I have done other GMATPrep questions where even you yourself have stated that they made a mistake
joondez
Sorry but this does not make sense.
"The product of the first 8 positive integers" cannot possible mean anything but 1*2*3*4*5*6*7*8.
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subhajeet
Given \(a, n > 1\), \(8! = a^n * K\)
so we have \(2^7 * 3^2 * 5 * 7 = a^n * K\).......(i)
Statement 1:
\(a^n = 64\), hence , we have \(a^n = 2^6\) or \(8^2\) or \(4^3\), therefore
\((2^6) * 2 * 3^2 * 5 * 7 = a^n * K\), \(a = 2, n = 6\)
\((8^2) * 2 * 3^2 * 5 * 7 = a^n * K\), \(a = 8 , n = 2\)
Hence, Statement 1 is Not Sufficient.
Statement 2:
\(n = 6\), hence we have from (i), that on factorization of \(8!\) the only integer greater than 1 & can accommodate \(6\) as its power is \(a = 2\).
Hence, Statement 2 is Sufficient.
Answer B.
Thanks,
GyM
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Expert reply
Posts: 22,722
dominion
Solution:
Statement One Only:
a^n = 64
Since a^n = 64, we see that a^n could be 2^6 or 4^4 or 8^2 or 64^1. Since a could be 2, 4, 8, or 64 (and n could be 6, 3, 2, or 1, respectively), we see that statement one alone is not sufficient.
Statement Two Only:
n = 6
It seems the statement is not sufficient; however, it is sufficient. That is because the product of the first 8 positive integers is:
8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 2 x 3 x 2^2 x 5 x 2 x 3 x 7 x 2^3 = 2^7 x 3^2 x 5 x 7
From the prime factorization of 8!, we see that the only integer that has an exponent that is at least 6 is 2, from the factor 2^7. We see that if 8! is a multiple of a^6 where a > 1, then a must be 2.
Answer: B
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If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?
The prime factorization of the product of the first 8 positive integers: 2^7 + 3^2 * 5 * 7
(1) a^n = 64
This tells us that a can be 2, 4, or 8, since 2^6 = 64, 4^3 = 64, 8^2 = 64. Insufficient.
(2) n=6
Lets look at the prime factorization of the product of the first 8 integers again: 2^7 + 3^2 * 5 * 7
Only 2 has a power of 6 or greater. Therefore a MUST be 6. Sufficient.
Answer is B.
The prime factorization of the product of the first 8 positive integers: 2^7 + 3^2 * 5 * 7
(1) a^n = 64
This tells us that a can be 2, 4, or 8, since 2^6 = 64, 4^3 = 64, 8^2 = 64. Insufficient.
(2) n=6
Lets look at the prime factorization of the product of the first 8 integers again: 2^7 + 3^2 * 5 * 7
Only 2 has a power of 6 or greater. Therefore a MUST be 6. Sufficient.
Answer is B.
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